Sum of the vertex degrees for vertices of degree greater than $1$ in a tree 
Let $T$ be a tree with $n_1$ vertices of degree 1 and $n_2$ other vertices. Prove that the sum of the vertex degrees for the vertices of degree greater than 1 is $$n_1+2(n_2−1).$$

I'm pretty new to graph theory and am very stuck on where to even start with this question. 
Any insight and help would be greatly appreciated.
 A: Hint:
The sum of the vertex degrees for ANY tree on $n$ vertices is always the same.  Do you know what it is?
If so, you can write $n=n_1+n_2$, plug it into the expression for the total vertex degree, and rearrange the equation to get what you need.
If you don't know the result for vertex degrees: do you know that the number of edges in a tree on $n$ vertices is always the same?  If you do, then you just need to realize that each edge contributes exactly two to the total vertex degree -- one for each vertex it connects to.
Update
You know that the total degree of the graph is $2(n-1)$. Therefore, we can write
$$
\begin{align*}
2(n-1)&=\sum_{v\in V}\deg(v)\\
&=\sum_{v\in V_1}\deg(v)+\sum_{v\in V_2}\deg(v),
\end{align*}
$$
where $V_1$ is the set of vertices of degree $1$, and $V_2$ is the set of vertices of degree $2$ or more.
Now, we know that
$$
\sum_{v\in V_1}\deg(v)=n_1,
$$
since there are $n_1$ such vertices and each has degree $1$. So, this tells us that
$$
\sum_{v\in V_2}\deg(v)=2(n-1)-n_1.
$$
But, we also know that $n_1+n_2=n$.  Can you see how to use this to finish up?
A: The sum of the degree of the vertices on a tree with $n$ nodes is $2(n-1)$, since there are $n-1$ edges. So we have
$$
\sum_{v} \deg v = 2(n-1)
$$
We can split up vertices into those that have degree $1$, and those that have degree greater than $1$. (We may assume the tree has at least two vertices, so there are no vertices of degree $0$.) So we get
$$
\sum_{v \;:\; \deg v = 1} \deg v
+ \sum_{v \;:\; \deg v > 1} \deg v
= 2(n-1)
$$
i.e.
$$
\sum_{v \;:\; \deg v = 1} 1
+ \sum_{v \;:\; \deg v > 1} \deg v
= 2(n-1).
$$
How many vertices are there with $\deg v = 1$? We are given that there are $n_1$ of them.
Also, convince yourself that $n = n_1 + n_2$. Plug both of these things in above and we get
$$
n_1
+ \sum_{v \;:\; \deg v > 1} \deg v
= 2(n_1 + n_2 -1),
$$
and rearranging you get the result you want.
